Diode Circuits: Part 1
M. B. Patil
mbpatil@ee.iitb.ac.in
www.ee.iitb.ac.in/~sequel
Department of Electrical Engineering
Indian Institute of Technology Bombay
M. B. Patil, IIT Bombay
Diodes
i
flow
V
pressure
M. B. Patil, IIT Bombay
Diodes
i
flow
V
pressure
* A diode may be thought of as the electrical counterpart of a directional valve (“check valve”).
M. B. Patil, IIT Bombay
Diodes
i
flow
V
pressure
* A diode may be thought of as the electrical counterpart of a directional valve (“check valve”).
* A check valve presents a small resistance if the pressure p > 0, but blocks the flow (i.e., presents a large
resistance) if p < 0.
M. B. Patil, IIT Bombay
Diodes
i
flow
V
pressure
* A diode may be thought of as the electrical counterpart of a directional valve (“check valve”).
* A check valve presents a small resistance if the pressure p > 0, but blocks the flow (i.e., presents a large
resistance) if p < 0.
* Similarly, a diode presents a small resistance in the forward direction and a large resistance in the reverse
direction.
M. B. Patil, IIT Bombay
Diodes
i
flow
V
pressure
* A diode may be thought of as the electrical counterpart of a directional valve (“check valve”).
* A check valve presents a small resistance if the pressure p > 0, but blocks the flow (i.e., presents a large
resistance) if p < 0.
* Similarly, a diode presents a small resistance in the forward direction and a large resistance in the reverse
direction.
* Note: In a practical diode, the resistance RD = V /i is a nonlinear function of the applied voltage V .
However, it is often a good approximation to treat it as a constant resistance which is small if V is
positive and large if V is negative.
M. B. Patil, IIT Bombay
Simple models: Ron /Roff model
i
i
V
R = Ron if V > 0
V
R = Roff if V < 0
M. B. Patil, IIT Bombay
Simple models: Ron /Roff model
i
i
V
R = Ron if V > 0
V
R = Roff if V < 0
* Since the resistance is different in the forward and reverse directions, the i − V relationship is not a
straight line.
M. B. Patil, IIT Bombay
Simple models: Ron /Roff model
i
i
V
R = Ron if V > 0
V
R = Roff if V < 0
* Since the resistance is different in the forward and reverse directions, the i − V relationship is not a
straight line.
* Examples:
M. B. Patil, IIT Bombay
Simple models: Ron /Roff model
i
i
V
R = Ron if V > 0
R = Roff if V < 0
V
* Since the resistance is different in the forward and reverse directions, the i − V relationship is not a
straight line.
* Examples:
10
Ron = 0.1 Ω
Roff = 500 Ω
Roff = 1 MΩ
i (mA)
Ron = 5 Ω
0
−5
−4
−3
−2
V (Volts)
−1
0
1
−5
−4
−3
−2
−1
0
1
V (Volts)
M. B. Patil, IIT Bombay
Simple models: ideal switch
i
i
i
i
S
S closed, V > 0
i
V
S open, V < 0
V
V
V
M. B. Patil, IIT Bombay
Simple models: ideal switch
i
i
i
i
S
S closed, V > 0
i
V
S open, V < 0
V
V
V
* Forward bias: i > 0 A, V = 0 V, → S is closed (a perfect contact).
M. B. Patil, IIT Bombay
Simple models: ideal switch
i
i
i
i
S
S closed, V > 0
i
V
S open, V < 0
V
V
V
* Forward bias: i > 0 A, V = 0 V, → S is closed (a perfect contact).
* Reverse bias: V < 0 V, i = 0 A → S is open (a perfect open circuit). The diode is said to “block” the
reverse applied voltage.
M. B. Patil, IIT Bombay
Simple models: ideal switch
i
i
i
i
S
S closed, V > 0
i
V
S open, V < 0
V
V
V
* Forward bias: i > 0 A, V = 0 V, → S is closed (a perfect contact).
* Reverse bias: V < 0 V, i = 0 A → S is open (a perfect open circuit). The diode is said to “block” the
reverse applied voltage.
* The actual values of V and i for a diode in a circuit get determined by the i-V relationship of the
diode and the constraints on V and i imposed by the circuit.
M. B. Patil, IIT Bombay
Shockley diode equation
i
i
p
V
n
V
M. B. Patil, IIT Bombay
Shockley diode equation
V
i = Is exp
− 1 , where VT = kB T /q .
VT
i
kB = Boltzmann’s constant = 1.38 × 10−23 J/K .
i
p
V
n
V
q = electron charge = 1.602 × 10−19 Coul.
T = temperature in ◦ K .
VT ≈ 25 mV at room temperature (27 ◦ C).
M. B. Patil, IIT Bombay
Shockley diode equation
V
i = Is exp
− 1 , where VT = kB T /q .
VT
i
kB = Boltzmann’s constant = 1.38 × 10−23 J/K .
i
p
V
n
V
q = electron charge = 1.602 × 10−19 Coul.
T = temperature in ◦ K .
VT ≈ 25 mV at room temperature (27 ◦ C).
* Is is called the “reverse saturation current.”
M. B. Patil, IIT Bombay
Shockley diode equation
V
i = Is exp
− 1 , where VT = kB T /q .
VT
i
kB = Boltzmann’s constant = 1.38 × 10−23 J/K .
i
p
V
n
V
q = electron charge = 1.602 × 10−19 Coul.
T = temperature in ◦ K .
VT ≈ 25 mV at room temperature (27 ◦ C).
* Is is called the “reverse saturation current.”
* For a typical low-power silicon diode, Is is of the order of 10−13 A (i.e., 0.1 pA).
M. B. Patil, IIT Bombay
Shockley diode equation
V
i = Is exp
− 1 , where VT = kB T /q .
VT
i
kB = Boltzmann’s constant = 1.38 × 10−23 J/K .
i
p
V
n
V
q = electron charge = 1.602 × 10−19 Coul.
T = temperature in ◦ K .
VT ≈ 25 mV at room temperature (27 ◦ C).
* Is is called the “reverse saturation current.”
* For a typical low-power silicon diode, Is is of the order of 10−13 A (i.e., 0.1 pA).
* Although Is is very small, it gets multiplied by a large exponential factor, giving a diode current of several
mA for V ≈ 0.7 V in a silicon diode.
M. B. Patil, IIT Bombay
Shockley diode equation
V
i = Is exp
− 1 , where VT = kB T /q .
VT
i
kB = Boltzmann’s constant = 1.38 × 10−23 J/K .
i
p
V
n
V
q = electron charge = 1.602 × 10−19 Coul.
T = temperature in ◦ K .
VT ≈ 25 mV at room temperature (27 ◦ C).
* Is is called the “reverse saturation current.”
* For a typical low-power silicon diode, Is is of the order of 10−13 A (i.e., 0.1 pA).
* Although Is is very small, it gets multiplied by a large exponential factor, giving a diode current of several
mA for V ≈ 0.7 V in a silicon diode.
* The “turn-on” voltage (Von ) of a diode depends on the value of Is . Von may be defined as the voltage at
which the diode starts carrying a substantial forward current (say, a few mA).
For a silicon diode, Von ≈ 0.7 V .
For LEDs, Von varies from about 1.8 V (red) to 3.3 V (blue).
M. B. Patil, IIT Bombay
Shockley diode equation
i
i
p
V
n
V
V
i = Is exp
− 1 , where VT = kB T /q .
VT
Example: Is = 1 × 10−13 A, VT = 25 mV .
M. B. Patil, IIT Bombay
Shockley diode equation
i
i
p
V
V
i = Is exp
− 1 , where VT = kB T /q .
VT
n
V
Example: Is = 1 × 10−13 A, VT = 25 mV .
V
x = V /VT
ex
i (Amp)
0.1
3.87
0.479×102
0.469×10−11
7.74
0.229×104
0.229×10−9
11.6
0.110×106
0.110×10−7
15.5
0.525×107
0.525×10−6
19.3
0.251×109
0.251×10−4
23.2
0.120×1011
0.120×10−2
24.0
0.260×1011
0.260×10−2
0.64
24.8
0.565×1011
0.565×10−2
0.66
25.5
0.122×1012
0.122×10−1
0.68
26.3
0.265×1012
0.265×10−1
0.70
27.1
0.575×1012
0.575×10−1
27.8
0.125×1013
0.125
0.2
0.3
0.4
0.5
0.6
0.62
0.72
M. B. Patil, IIT Bombay
Shockley diode equation
i
V
V
i = Is exp
− 1 , where VT = kB T /q .
VT
n
V
Example: Is = 1 × 10−13 A, VT = 25 mV .
100
V
x = V /VT
ex
i (Amp)
0.1
3.87
0.479×102
0.469×10−11
7.74
0.229×104
0.229×10−9
0.3
11.6
0.110×106
0.110×10−7
0.4
15.5
0.525×107
0.525×10−6
19.3
0.251×109
0.251×10−4
0.6
23.2
0.120×1011
0.120×10−2
0.62
24.0
0.260×1011
0.260×10−2
0.64
24.8
0.565×1011
0.565×10−2
0.66
25.5
0.122×1012
0.122×10−1
0.68
26.3
0.265×1012
0.265×10−1
0.70
27.1
0.575×1012
0.575×10−1
0.72
27.8
0.125×1013
0.125
0.2
0.5
i (Amp)
p
log scale
10−6
10−12
100
linear scale
80
i (mA)
i
60
40
20
0
0
0.2
0.4
V (Volts)
0.6
0.8
M. B. Patil, IIT Bombay
Diode circuit model
100
80
i (mA)
i
60
i
V > Von :
Ron
V
40
slope = 1/Ron
actual
model
20
0
Von
0
0.2
0.4
V (Volts)
0.6
V
V < Von :
0.8
Von
* In many circuits, Ron can be neglected (assumed to be 0 Ω) since it is much smaller than the other
resistances in the circuit. In that case, the diode in forward conduction can be replaced with simply a
battery.
M. B. Patil, IIT Bombay
Diode circuit model
100
80
i (mA)
i
60
i
V > Von :
Ron
V
40
slope = 1/Ron
actual
model
20
0
Von
0
0.2
0.4
V (Volts)
0.6
V
V < Von :
0.8
Von
* In many circuits, Ron can be neglected (assumed to be 0 Ω) since it is much smaller than the other
resistances in the circuit. In that case, the diode in forward conduction can be replaced with simply a
battery.
* Note that the “battery” shown in the above model is not a “source” of power! It can only absorb power
(see the direction of the current), causing heat dissipation.
M. B. Patil, IIT Bombay
Diode circuit analysis
* In DC situations, for each diode in the circuit, we need to establish whether it is on or off, replace it with
the corresponding equivalent circuit, and then obtain the quantities of interest.
M. B. Patil, IIT Bombay
Diode circuit analysis
* In DC situations, for each diode in the circuit, we need to establish whether it is on or off, replace it with
the corresponding equivalent circuit, and then obtain the quantities of interest.
* In transient analysis, we need to find the time points at which a diode turns on or off, and analyse the
circuit in intervals between these time points.
M. B. Patil, IIT Bombay
Diode circuit analysis
* In DC situations, for each diode in the circuit, we need to establish whether it is on or off, replace it with
the corresponding equivalent circuit, and then obtain the quantities of interest.
* In transient analysis, we need to find the time points at which a diode turns on or off, and analyse the
circuit in intervals between these time points.
* In some diode circuits, the exponential nature of the diode I-V relationship (the Shockley model) is made
use of. For these circuits, computation is usually difficult, and computer simulation may be required to
solve the resulting non-linear equations.
M. B. Patil, IIT Bombay
Diode circuit example
6k
A
B
R1
D
R2
36 V
3k
C
i=?
ID
R3
1k
0.7 V
VD
Diode circuit example
6k
A
B
R1
i=?
D
R3
R2
36 V
1k
3k
0.7 V
C
Case 1: D is off.
6k
A
B
R1
i
R2
36 V
3k
C
ID
R3
1k
VD
Diode circuit example
6k
A
B
R1
i=?
D
R3
R2
36 V
1k
3k
0.7 V
C
Case 1: D is off.
6k
A
B
R1
i
R2
36 V
3k
R3
1k
C
3
× 36 = 12 V ,
9
which is not consistent with our
assumption of D being off.
VAB = VAC =
ID
VD
Diode circuit example
6k
A
B
R1
i=?
D
R3
R2
36 V
1k
3k
0.7 V
C
Case 1: D is off.
6k
A
B
R1
i
R2
36 V
ID
3k
R3
1k
C
3
× 36 = 12 V ,
9
which is not consistent with our
assumption of D being off.
VAB = VAC =
→ D must be on.
VD
Diode circuit example
6k
A
B
R1
i=?
D
R3
R2
36 V
1k
3k
Case 2: D is on.
Case 1: D is off.
A
6k
B
R1
i
R2
36 V
VD
0.7 V
C
6k
ID
3k
R3
1k
C
3
× 36 = 12 V ,
9
which is not consistent with our
assumption of D being off.
VAB = VAC =
→ D must be on.
0.7 V
A
B
i
R1
R2
36 V
3k
C
R3
1k
Diode circuit example
6k
A
B
R1
i=?
D
R3
R2
36 V
1k
3k
Case 2: D is on.
Case 1: D is off.
A
6k
B
R1
i
R2
36 V
VD
0.7 V
C
6k
ID
3k
R3
1k
C
3
× 36 = 12 V ,
9
which is not consistent with our
assumption of D being off.
VAB = VAC =
→ D must be on.
0.7 V
A
B
i
R1
R2
36 V
3k
R3
1k
C
Taking VC = 0 V,
VA − 0.7
VA − 36 VA
+
+
= 0,
6k
3k
1k
→ VA = 4.47 V, i = 3.77 mA .
Diode circuit example
6k
A
B
R1
i=?
D
R3
R2
36 V
1k
3k
Case 2: D is on.
Case 1: D is off.
A
6k
B
R1
i
R2
36 V
VD
0.7 V
C
6k
ID
3k
R3
1k
C
3
× 36 = 12 V ,
9
which is not consistent with our
assumption of D being off.
VAB = VAC =
→ D must be on.
0.7 V
A
B
i
R1
R2
36 V
3k
R3
1k
C
Taking VC = 0 V,
VA − 0.7
VA − 36 VA
+
+
= 0,
6k
3k
1k
→ VA = 4.47 V, i = 3.77 mA .
Remark: Often, we can figure out by inspection if a diode is on or off.
M. B. Patil, IIT Bombay
Diode circuit example
R
A
ID
1k
D1
D2
1V
Vi
R2
Vo
0.7 V
VD
0.5 k
(a) Plot Vo versus Vi for −5 V < Vi < 5 V .
R1
1.5 k
i1
B
i2
(b) Plot Vo (t) for a triangular input:
−5 V to +5 V, 500 Hz .
M. B. Patil, IIT Bombay
Diode circuit example
R
A
ID
1k
D1
D2
1V
Vi
R2
0.7 V
Vo
VD
0.5 k
(a) Plot Vo versus Vi for −5 V < Vi < 5 V .
R1
1.5 k
i1
B
i2
(b) Plot Vo (t) for a triangular input:
−5 V to +5 V, 500 Hz .
First, let us show that D1 on ⇒ D2 off, and D2 on ⇒ D1 off.
M. B. Patil, IIT Bombay
Diode circuit example
R
A
ID
1k
D1
D2
1V
Vi
R2
0.7 V
Vo
VD
0.5 k
(a) Plot Vo versus Vi for −5 V < Vi < 5 V .
R1
1.5 k
i1
i2
B
(b) Plot Vo (t) for a triangular input:
−5 V to +5 V, 500 Hz .
First, let us show that D1 on ⇒ D2 off, and D2 on ⇒ D1 off.
Consider D1 to be on → VAB = 0.7 + 1 + i1 R1 .
M. B. Patil, IIT Bombay
Diode circuit example
R
A
ID
1k
D1
D2
1V
Vi
R2
0.7 V
Vo
VD
0.5 k
(a) Plot Vo versus Vi for −5 V < Vi < 5 V .
R1
1.5 k
i1
i2
B
(b) Plot Vo (t) for a triangular input:
−5 V to +5 V, 500 Hz .
First, let us show that D1 on ⇒ D2 off, and D2 on ⇒ D1 off.
Consider D1 to be on → VAB = 0.7 + 1 + i1 R1 .
Note that i1 > 0, since D1 can only conduct in the forward direction.
⇒ VAB > 1.7 V ⇒ D2 cannot conduct.
M. B. Patil, IIT Bombay
Diode circuit example
R
A
ID
1k
D1
D2
1V
Vi
R2
0.7 V
Vo
VD
0.5 k
(a) Plot Vo versus Vi for −5 V < Vi < 5 V .
R1
1.5 k
i1
i2
B
(b) Plot Vo (t) for a triangular input:
−5 V to +5 V, 500 Hz .
First, let us show that D1 on ⇒ D2 off, and D2 on ⇒ D1 off.
Consider D1 to be on → VAB = 0.7 + 1 + i1 R1 .
Note that i1 > 0, since D1 can only conduct in the forward direction.
⇒ VAB > 1.7 V ⇒ D2 cannot conduct.
Similarly, if D2 is on, VBA > 0.7 V , i.e., VAB < −0.7 V ⇒ D1 cannot conduct.
M. B. Patil, IIT Bombay
Diode circuit example
R
A
ID
1k
D1
D2
1V
Vi
R2
0.7 V
Vo
VD
0.5 k
(a) Plot Vo versus Vi for −5 V < Vi < 5 V .
R1
1.5 k
i1
i2
B
(b) Plot Vo (t) for a triangular input:
−5 V to +5 V, 500 Hz .
First, let us show that D1 on ⇒ D2 off, and D2 on ⇒ D1 off.
Consider D1 to be on → VAB = 0.7 + 1 + i1 R1 .
Note that i1 > 0, since D1 can only conduct in the forward direction.
⇒ VAB > 1.7 V ⇒ D2 cannot conduct.
Similarly, if D2 is on, VBA > 0.7 V , i.e., VAB < −0.7 V ⇒ D1 cannot conduct.
Clearly, D1 on ⇒ D2 off, and D2 on ⇒ D1 off.
M. B. Patil, IIT Bombay
Diode circuit example
R
A
D1
D2
1V
Vi
Vo
R2
R1
i1
B
i2
Diode circuit example
R
A
D1
D2
1V
Vi
Vo
R2
R1
i1
i2
B
D1 on:
Vi = i1 (R + R1 ) + 1 + 0.7
Diode circuit example
R
A
D1
D2
1V
Vi
Vo
R2
R1
i1
i2
B
D1 on:
Vi = i1 (R + R1 ) + 1 + 0.7
Since i1 > 0, Vi > 1.7 V
Diode circuit example
R
R
A
D1
D1
D2
1V
Vi
A
Vo
1V
Vi
R2
Vo
R2
R1
i1
D2
R1
i2
B
D1 on:
Vi = i1 (R + R1 ) + 1 + 0.7
Since i1 > 0, Vi > 1.7 V
i2
B
Diode circuit example
R
R
A
D1
D1
D2
1V
Vi
A
Vo
1V
Vi
R2
Vo
R2
R1
i1
D2
R1
i2
i2
B
D1 on:
Vi = i1 (R + R1 ) + 1 + 0.7
Since i1 > 0, Vi > 1.7 V
B
D2 on:
i2 (R + R2 ) + 0.7 + Vi = 0
Diode circuit example
R
R
A
D1
D1
D2
1V
Vi
A
Vo
1V
Vi
R2
Vo
R2
R1
i1
D2
R1
i2
i2
B
D1 on:
B
D2 on:
Vi = i1 (R + R1 ) + 1 + 0.7
i2 (R + R2 ) + 0.7 + Vi = 0
Since i1 > 0, Vi > 1.7 V
Vi = − [0.7 + i2 (R + R2 )]
Diode circuit example
R
R
A
D1
D1
D2
1V
Vi
A
Vo
1V
Vi
R2
Vo
R2
R1
i1
D2
R1
i2
i2
B
D1 on:
B
D2 on:
Vi = i1 (R + R1 ) + 1 + 0.7
i2 (R + R2 ) + 0.7 + Vi = 0
Since i1 > 0, Vi > 1.7 V
Vi = − [0.7 + i2 (R + R2 )]
Since i2 > 0, Vi < −0.7 V
Diode circuit example
R
R
A
D1
D1
D2
1V
Vi
A
Vo
1V
Vi
R2
Vo
R2
R1
i1
D2
R1
i2
i2
B
D1 on:
B
D2 on:
Vi = i1 (R + R1 ) + 1 + 0.7
i2 (R + R2 ) + 0.7 + Vi = 0
Since i1 > 0, Vi > 1.7 V
Vi = − [0.7 + i2 (R + R2 )]
Since i2 > 0, Vi < −0.7 V
For −0.7 V < Vi < 1.7 V, neither D1 nor D2 can conduct.
M. B. Patil, IIT Bombay
R
* For −0.7 V < Vi < 1.7 V , both D1 and D2 are off.
→ no drop across R, and Vo = Vi .
(1)
A
1k
D1
D2
1V
Vi
Vo
R2
0.5 k
R1
1.5 k
i1
i2
B
M. B. Patil, IIT Bombay
R
* For −0.7 V < Vi < 1.7 V , both D1 and D2 are off.
→ no drop across R, and Vo = Vi .
(1)
A
1k
D1
D2
1V
Vi
Vo
R2
0.5 k
R1
1.5 k
i2
i1
B
Vo
5
0
D1 off
D2 on
D1 off
D2 off
−5
−5
0
−0.7
D1 on
D2 off
5
Vi
1.7
M. B. Patil, IIT Bombay
R
* For −0.7 V < Vi < 1.7 V , both D1 and D2 are off.
→ no drop across R, and Vo = Vi .
(1)
* For Vi < −0.7 V , D2 conducts. → Vo = −0.7 − i2 R2 .
Use KVL to get i2 : Vi + i2 R2 + 0.7 + Ri2 = 0.
Vi + 0.7
→ i2 = −
, and
R + R2
R
R2
Vi − 0.7
.
(2)
Vo = −0.7 − R2 i2 =
R + R2
R + R2
dVo
R2
0.5 k
1
Slope
=
=
= .
dVi
R + R2
1 k + 0.5 k
3
A
1k
D1
D2
1V
Vi
Vo
R2
0.5 k
R1
1.5 k
i2
i1
B
Vo
5
0
D1 off
D2 on
D1 off
D2 off
−5
−5
0
−0.7
D1 on
D2 off
5
Vi
1.7
M. B. Patil, IIT Bombay
R
* For −0.7 V < Vi < 1.7 V , both D1 and D2 are off.
→ no drop across R, and Vo = Vi .
(1)
* For Vi < −0.7 V , D2 conducts. → Vo = −0.7 − i2 R2 .
Use KVL to get i2 : Vi + i2 R2 + 0.7 + Ri2 = 0.
Vi + 0.7
→ i2 = −
, and
R + R2
R
R2
Vi − 0.7
.
(2)
Vo = −0.7 − R2 i2 =
R + R2
R + R2
dVo
R2
0.5 k
1
Slope
=
=
= .
dVi
R + R2
1 k + 0.5 k
3
* For Vi > 1.7 V , D1 conducts. → Vo = 0.7 + 1 + i1 R1 .
Use KVL to get i1 : −Vi + i1 R + 0.7 + 1 + i1 R1 = 0.
Vi − 1.7
, and
→ i1 =
R + R1
R1
R
Vo = 1.7 + R1 i1 =
Vi + 1.7
.
(3)
R + R1
R + R1
dVo
R1
1.5 k
3
Slope
=
=
= .
dVi
R + R1
1 k + 1.5 k
5
A
1k
D1
D2
1V
Vi
Vo
R2
0.5 k
R1
1.5 k
i2
i1
B
Vo
5
0
D1 off
D2 on
D1 off
D2 off
−5
−5
0
−0.7
D1 on
D2 off
5
Vi
1.7
M. B. Patil, IIT Bombay
R
* For −0.7 V < Vi < 1.7 V , both D1 and D2 are off.
→ no drop across R, and Vo = Vi .
(1)
* For Vi < −0.7 V , D2 conducts. → Vo = −0.7 − i2 R2 .
Use KVL to get i2 : Vi + i2 R2 + 0.7 + Ri2 = 0.
Vi + 0.7
→ i2 = −
, and
R + R2
R
R2
Vi − 0.7
.
(2)
Vo = −0.7 − R2 i2 =
R + R2
R + R2
dVo
R2
0.5 k
1
Slope
=
=
= .
dVi
R + R2
1 k + 0.5 k
3
* For Vi > 1.7 V , D1 conducts. → Vo = 0.7 + 1 + i1 R1 .
Use KVL to get i1 : −Vi + i1 R + 0.7 + 1 + i1 R1 = 0.
Vi − 1.7
, and
→ i1 =
R + R1
R1
R
Vo = 1.7 + R1 i1 =
Vi + 1.7
.
(3)
R + R1
R + R1
dVo
R1
1.5 k
3
Slope
=
=
= .
dVi
R + R1
1 k + 1.5 k
5
A
1k
D1
D2
1V
Vi
Vo
R2
0.5 k
R1
1.5 k
i2
i1
B
Vo
5
0
* Using Eqs. (1)-(3), we plot Vo versus Vi .
D1 off
(SEQUEL file: ee101 diode circuit 1.sqproj)
D2 on
D1 off
D2 off
−5
−5
0
−0.7
D1 on
D2 off
5
Vi
1.7
M. B. Patil, IIT Bombay
R
A
Vo
D1
Vi
D2
1V
R2
Vo
Vo
5
1k
D1 off
D1 off
D1 on
D2 on
D2 off
D2 off
5
0
0
0.5 k
R1
1.5 k
i1
i2
B
0
5
−5
0
5
Vi
−5
t1
1 t2
2 t (msec)
1
2
0
Vi
t (msec)
0.5
Two time points, t1 and t2,
are shown as examples.
0
0
0
Point-by-point construction of
Vo versus t:
−5
−5
1
1
1.5
t (msec)
t (msec)
2
2
M. B. Patil, IIT Bombay
Diode circuit example
D
1k
ID
R1
Vi
R2
0.5 k
Vo
0.7 V
Plot Vo versus Vi for −5 V < Vi < 5 V .
VD
Diode circuit example
D
0.7 V
1k
ID
R1
Vi
R2
0.5 k
R1
Vo
Vi
0.7 V
Plot Vo versus Vi for −5 V < Vi < 5 V .
R2
VD
D on
Vo = Vi − 0.7
Vo
Diode circuit example
D
VD
0.7 V
1k
ID
R1
Vi
R2
0.5 k
R1
Vo
Vi
0.7 V
Plot Vo versus Vi for −5 V < Vi < 5 V .
R1
R2
Vo
R2
Vi
Vo
VD
D off
D on
Vo = Vi − 0.7
Vo =
R2
Vi
R1 + R2
M. B. Patil, IIT Bombay
Diode circuit example
D
VD
0.7 V
1k
ID
R1
Vi
R2
0.5 k
R1
Vo
Vi
0.7 V
R1
R2
Vo
R2
Vi
Vo
VD
Plot Vo versus Vi for −5 V < Vi < 5 V .
D off
D on
Vo = Vi − 0.7
Vo =
R2
Vi
R1 + R2
At what value of Vi will the diode turn on?
M. B. Patil, IIT Bombay
Diode circuit example
D
VD
0.7 V
1k
ID
R1
Vi
R2
0.5 k
R1
Vo
Vi
0.7 V
R1
R2
Vo
R2
Vi
Vo
VD
Plot Vo versus Vi for −5 V < Vi < 5 V .
D off
D on
Vo = Vi − 0.7
Vo =
R2
Vi
R1 + R2
At what value of Vi will the diode turn on?
R1
In the off state, VD =
Vi .
R1 + R2
M. B. Patil, IIT Bombay
Diode circuit example
D
VD
0.7 V
1k
ID
R1
Vi
R2
0.5 k
R1
Vo
Vi
0.7 V
R1
R2
Vo
R2
Vi
Vo
VD
Plot Vo versus Vi for −5 V < Vi < 5 V .
D off
D on
Vo = Vi − 0.7
Vo =
R2
Vi
R1 + R2
At what value of Vi will the diode turn on?
R1
In the off state, VD =
Vi .
R1 + R2
As Vi increases, VD increases.
For D to turn on, we need VD = 0.7 V .
R1 + R2
i.e., Vi =
× 0.7 = 1.05 V .
R1
M. B. Patil, IIT Bombay
Diode circuit example
D
VD
0.7 V
1k
ID
R1
Vi
R2
0.5 k
R1
Vo
Vi
0.7 V
R1
R2
Vo
R2
Vi
Vo
VD
Plot Vo versus Vi for −5 V < Vi < 5 V .
D off
D on
Vo = Vi − 0.7
Vo =
R2
Vi
R1 + R2
At what value of Vi will the diode turn on?
R1
In the off state, VD =
Vi .
R1 + R2
As Vi increases, VD increases.
For D to turn on, we need VD = 0.7 V .
R1 + R2
i.e., Vi =
× 0.7 = 1.05 V .
R1
(SEQUEL file: ee101 diode circuit 2.sqproj)
M. B. Patil, IIT Bombay
Diode circuit example
D
VD
0.7 V
1k
ID
R1
Vi
R2
0.5 k
R1
Vo
Vi
0.7 V
R1
R2
Vo
R2
Vi
Vo
VD
Plot Vo versus Vi for −5 V < Vi < 5 V .
Vo
At what value of Vi will the diode turn on?
R1
In the off state, VD =
Vi .
R1 + R2
As Vi increases, VD increases.
D off
D on
Vo = Vi − 0.7
Vo =
R2
Vi
R1 + R2
5
0
For D to turn on, we need VD = 0.7 V .
R1 + R2
i.e., Vi =
× 0.7 = 1.05 V .
R1
(SEQUEL file: ee101 diode circuit 2.sqproj)
D off
−5
−5
D on
0
5
Vi
1.05
M. B. Patil, IIT Bombay
Diode circuit example
1V
R1
1k
Plot Vo versus Vi (Ref: Sedra/Smith).
ID
D1
Vi
R
Vo
3k
D2
D3
R2
1k
−2 V
0.7 V
VD
Diode circuit example
1V
R1
1k
Plot Vo versus Vi (Ref: Sedra/Smith).
ID
D1
Vi
R
Vo
3k
D2
D3
R2
1k
−2 V
0.7 V
VD
It is easier to find the status (on/off) of each diode w. r. t. Vo .
Diode circuit example
1V
Plot Vo versus Vi (Ref: Sedra/Smith).
R1
1k
ID
D1
Vi
R
Vo
3k
D2
D3
VD
0.7 V
It is easier to find the status (on/off) of each diode w. r. t. Vo .
R2
1k
−2 V
−4
−3
−2
−1
0
1
2
Vo
Diode circuit example
1V
Plot Vo versus Vi (Ref: Sedra/Smith).
R1
1k
ID
D1
Vi
R
Vo
3k
D2
D3
VD
0.7 V
It is easier to find the status (on/off) of each diode w. r. t. Vo .
R2
1k
−2 V
D1 on
−4
−3
−2
−1
0
1
2
Vo
Diode circuit example
1V
Plot Vo versus Vi (Ref: Sedra/Smith).
R1
1k
ID
D1
Vi
R
Vo
3k
D2
D3
It is easier to find the status (on/off) of each diode w. r. t. Vo .
R2
1k
−2 V
VD
0.7 V
D2
on
−4
−3
D1 on
−2
−1
0
1
2
Vo
Diode circuit example
1V
Plot Vo versus Vi (Ref: Sedra/Smith).
R1
1k
ID
D1
Vi
R
Vo
3k
D2
D3
R2
1k
−2 V
VD
0.7 V
It is easier to find the status (on/off) of each diode w. r. t. Vo .
D2 and D3 on
−4
D2
on
−3
D1 on
−2
−1
0
1
2
Vo
M. B. Patil, IIT Bombay
1V
R1
1k
D1
i
Vi
Vo
3k R
D2
D3
R2
1k
−2 V
D2 and D3 on
−4
D2
on
−3
D1 on
−2
−1
0
1
2
Vo
1V
R1
1k
D1
i
Vi
D2 and D3 on
−4
D2
on
−3
D1 on
−2
−1
When D1 just starts conducting,
Vo
3k R
D2
D3
R2
1k
−2 V
Vo = 1.7 V, i ≈ 0 → Vi = 1.7 V
0
1
2
Vo
1V
R1
1k
D1
i
Vi
D2 and D3 on
−4
D2
on
−3
D1 on
−2
−1
0
When D1 just starts conducting,
Vo
3k R
D2
D3
R2
1k
−2 V
Vo = 1.7 V, i ≈ 0 → Vi = 1.7 V
For Vi > 1.7 V, Vo = 1.7 +
!
Vi − 1.7
R1
R + R1
1
2
Vo
1V
R1
1k
D1
i
Vi
D2 and D3 on
−4
D2
on
−3
D1 on
−2
−1
0
When D1 just starts conducting,
Vo
3k R
D2
Vo = 1.7 V, i ≈ 0 → Vi = 1.7 V
For Vi > 1.7 V, Vo = 1.7 +
Slope
D3
R2
1k
−2 V
dVo
R1
1
=
=
dVi
R + R1
4
!
Vi − 1.7
R1
R + R1
1
2
Vo
1V
R1
1k
D1
i
D2
on
−4
D1 on
−3
−2
−1
0
1
2
Vo
When D1 just starts conducting,
Vo
3k R
D2
Vo = 1.7 V, i ≈ 0 → Vi = 1.7 V
For Vi > 1.7 V, Vo = 1.7 +
Slope
D3
R2
1k
−2 V
dVo
R1
1
=
=
dVi
R + R1
4
!
Vi − 1.7
R1
R + R1
4
2
Vo (Volts)
Vi
D2 and D3 on
0
D1 on
−2
−4
−8
−4
0
4
8
Vi (Volts)
M. B. Patil, IIT Bombay
1V
R1
1k
D1
Vi
3k R
Vo
i
D2
D3
R2
1k
−2 V
D2 and D3 on
−4
D2
on
−3
D1 on
−2
−1
0
1
2
Vo
1V
R1
1k
D1
Vi
3k R
Vo
i
D2
D3
R2
1k
−2 V
D2 and D3 on
−4
D2
on
−3
D1 on
−2
−1
0
When D2 just starts conducting,
Vo = −2.7 V, i ≈ 0 → Vi = −2.7 V
1
2
Vo
1V
R1
1k
D1
Vi
3k R
Vo
i
D2
D3
R2
1k
−2 V
D2 and D3 on
−4
D2
on
−3
D1 on
−2
−1
0
1
When D2 just starts conducting,
Vo = −2.7 V, i ≈ 0 → Vi = −2.7 V
For Vi < −2.7 V, Vo = Vi +
!
−2.7 − Vi
R
R + R2
2
Vo
1V
R1
1k
D1
Vi
3k R
Vo
i
D2
D2 and D3 on
−4
−3
D3
−2 V
D1 on
−2
−1
0
1
When D2 just starts conducting,
Vo = −2.7 V, i ≈ 0 → Vi = −2.7 V
For Vi < −2.7 V, Vo = Vi +
Slope
R2
1k
D2
on
!
−2.7 − Vi
R
R + R2
dVo
R
R2
1
=1−
=
=
dVi
R + R2
R + R2
4
2
Vo
1V
R1
1k
D1
D2
−3
−1
For Vi < −2.7 V, Vo = Vi +
R2
1k
−2 V
−2
0
Vo = −2.7 V, i ≈ 0 → Vi = −2.7 V
Slope
D3
D1 on
1
2
Vo
When D2 just starts conducting,
Vo
i
D2
on
−4
!
−2.7 − Vi
R
R + R2
dVo
R
R2
1
=1−
=
=
dVi
R + R2
R + R2
4
4
2
Vo (Volts)
Vi
3k R
D2 and D3 on
0
D2 on
D1 on
−2
−4
−8
−4
0
4
8
Vi (Volts)
M. B. Patil, IIT Bombay
1V
R1
1k
D1
Vi
3k R
Vo
i
D2
D3
R2
1k
−2 V
D2 and D3 on
−4
D2
on
−3
D1 on
−2
−1
0
1
2
Vo
1V
R1
1k
D1
Vi
3k R
Vo
i
D2
D3
R2
1k
−2 V
D2 and D3 on
−4
D2
on
−3
D1 on
−2
−1
0
When D3 just starts conducting,
R2
(−2.7 − Vi )
= 0.7 V → Vi = −5.5 V
R + R2
1
2
Vo
1V
R1
1k
D1
Vi
3k R
Vo
i
D2
D3
R2
1k
−2 V
D2 and D3 on
−4
D2
on
−3
D1 on
−2
−1
0
When D3 just starts conducting,
R2
(−2.7 − Vi )
= 0.7 V → Vi = −5.5 V
R + R2
Vo = −2 − 0.7 − 0.7 = −3.4 V
1
2
Vo
1V
R1
1k
D1
Vi
3k R
Vo
i
D2
D3
R2
1k
−2 V
D2 and D3 on
−4
D2
on
−3
D1 on
−2
−1
0
When D3 just starts conducting,
R2
(−2.7 − Vi )
= 0.7 V → Vi = −5.5 V
R + R2
Vo = −2 − 0.7 − 0.7 = −3.4 V
For Vi < −5.5 V, Vo = −3.4 V (constant)
1
2
Vo
1V
R1
1k
D1
D2
D3
R2
1k
−2 V
D1 on
−3
−2
−1
0
1
When D3 just starts conducting,
Vo
i
D2
on
−4
(−2.7 − Vi )
= 0.7 V → Vi = −5.5 V
R + R2
R2
Vo = −2 − 0.7 − 0.7 = −3.4 V
For Vi < −5.5 V, Vo = −3.4 V (constant)
4
2
Vo (Volts)
Vi
3k R
D2 and D3 on
0
−2
D2 on D2 on
D3 on
−4
−8
−4
D1 on
0
Vi (Volts)
4
8
2
Vo
1V
D2 and D3 on
R1
1k
D1
−4
Vo
i
D2
D3
R2
1k
−2 V
−3
ee101 diode circuit 12.sqproj
−2
−1
0
1
2
Vo
(−2.7 − Vi )
= 0.7 V → Vi = −5.5 V
R + R2
R2
Vo = −2 − 0.7 − 0.7 = −3.4 V
For Vi < −5.5 V, Vo = −3.4 V (constant)
4
2
0
−2
SEQUEL file:
D1 on
When D3 just starts conducting,
Vo (Volts)
Vi
3k R
D2
on
D2 on D2 on
D3 on
−4
−8
−4
D1 on
0
4
8
Vi (Volts)
M. B. Patil, IIT Bombay
Reverse breakdown
40
i
V
i (mA)
20
0
−20
−6
−5
−4
−3
−2
−1
0
1
V (Volts)
M. B. Patil, IIT Bombay
Reverse breakdown
40
i
V
i (mA)
20
0
−20
−6
−5
−4
−3
−2
−1
0
1
V (Volts)
* In the reverse direction, an ideal diode presents a large resistance for any applied voltage.
M. B. Patil, IIT Bombay
Reverse breakdown
40
i
V
i (mA)
20
0
−20
−6
−5
−4
−3
−2
−1
0
1
V (Volts)
* In the reverse direction, an ideal diode presents a large resistance for any applied voltage.
* A real diode cannot withstand indefinitely large reverse voltages and “breaks down” at a certain voltage
called the “breakdown voltage” (VBR ).
M. B. Patil, IIT Bombay
Reverse breakdown
40
i
V
i (mA)
20
0
−20
−6
−5
−4
−3
−2
−1
0
1
V (Volts)
* In the reverse direction, an ideal diode presents a large resistance for any applied voltage.
* A real diode cannot withstand indefinitely large reverse voltages and “breaks down” at a certain voltage
called the “breakdown voltage” (VBR ).
* When the reverse bias VR > VBR (i.e., V < −VBR ), the diode allows a large amount of current. If the
current is not constrained by the external circuit, the diode would get damaged.
M. B. Patil, IIT Bombay
Reverse breakdown
40
i
V
i (mA)
20
0
Symbol for a Zener diode
−20
−6
−5
−4
−3
−2
−1
0
1
V (Volts)
M. B. Patil, IIT Bombay
Reverse breakdown
40
i
V
i (mA)
20
0
Symbol for a Zener diode
−20
−6
−5
−4
−3
−2
−1
0
1
V (Volts)
* A wide variety of diodes is available, with VBR ranging from a few Volts to a few thousand Volts!
Generally, higher the breakdown voltage, higher is the cost.
M. B. Patil, IIT Bombay
Reverse breakdown
40
i
V
i (mA)
20
0
Symbol for a Zener diode
−20
−6
−5
−4
−3
−2
−1
0
1
V (Volts)
* A wide variety of diodes is available, with VBR ranging from a few Volts to a few thousand Volts!
Generally, higher the breakdown voltage, higher is the cost.
* Diodes with high VBR are generally used in power electronics applications and are therefore also designed
to carry a large forward current (tens or hundreds of Amps).
M. B. Patil, IIT Bombay
Reverse breakdown
40
i
V
i (mA)
20
0
Symbol for a Zener diode
−20
−6
−5
−4
−3
−2
−1
0
1
V (Volts)
* A wide variety of diodes is available, with VBR ranging from a few Volts to a few thousand Volts!
Generally, higher the breakdown voltage, higher is the cost.
* Diodes with high VBR are generally used in power electronics applications and are therefore also designed
to carry a large forward current (tens or hundreds of Amps).
* Typically, circuits are designed so that the reverse bias across any diode is less than the VBR rating for that
diode.
M. B. Patil, IIT Bombay
Reverse breakdown
40
i
V
i (mA)
20
0
Symbol for a Zener diode
−20
−6
−5
−4
−3
−2
−1
0
1
V (Volts)
* A wide variety of diodes is available, with VBR ranging from a few Volts to a few thousand Volts!
Generally, higher the breakdown voltage, higher is the cost.
* Diodes with high VBR are generally used in power electronics applications and are therefore also designed
to carry a large forward current (tens or hundreds of Amps).
* Typically, circuits are designed so that the reverse bias across any diode is less than the VBR rating for that
diode.
* “Zener” diodes typically have VBR of a few Volts, which is denoted by VZ . They are often used to limit
the voltage swing in electronic circuits.
M. B. Patil, IIT Bombay
Two Zener diodes connected “back-to-back”
i1 (or i2 )
V1
i1 D1
i
V2
D2 i2
V
−VZ
Von V1 (or V2 )
M. B. Patil, IIT Bombay
Two Zener diodes connected “back-to-back”
i1 (or i2 )
V1
i1 D1
i
V2
D2 i2
−VZ
V
Von V1 (or V2 )
* i > 0 → D1 in forward conduction, D2 in reverse conduction
M. B. Patil, IIT Bombay
Two Zener diodes connected “back-to-back”
i1 (or i2 )
V1
i1 D1
i
V2
D2 i2
−VZ
V
Von V1 (or V2 )
* i > 0 → D1 in forward conduction, D2 in reverse conduction
→ V1 = Von , V2 = −VZ .
M. B. Patil, IIT Bombay
Two Zener diodes connected “back-to-back”
i1 (or i2 )
V1
i1 D1
i
V2
D2 i2
−VZ
V
Von V1 (or V2 )
* i > 0 → D1 in forward conduction, D2 in reverse conduction
→ V1 = Von , V2 = −VZ .
Total voltage drop V = V1 − V2 = Von + VZ .
M. B. Patil, IIT Bombay
Two Zener diodes connected “back-to-back”
i1 (or i2 )
V1
i1 D1
i
V2
D2 i2
−VZ
V
Von V1 (or V2 )
* i > 0 → D1 in forward conduction, D2 in reverse conduction
→ V1 = Von , V2 = −VZ .
Total voltage drop V = V1 − V2 = Von + VZ .
Example: Von = 0.7 V, VZ = 5 V → V = 5.7 V.
M. B. Patil, IIT Bombay
Two Zener diodes connected “back-to-back”
i1 (or i2 )
V1
i1 D1
i
V2
D2 i2
−VZ
V
Von V1 (or V2 )
* i > 0 → D1 in forward conduction, D2 in reverse conduction
→ V1 = Von , V2 = −VZ .
Total voltage drop V = V1 − V2 = Von + VZ .
Example: Von = 0.7 V, VZ = 5 V → V = 5.7 V.
* i < 0 → D1 in reverse conduction, D2 in forward conduction
M. B. Patil, IIT Bombay
Two Zener diodes connected “back-to-back”
i1 (or i2 )
V1
i1 D1
i
V2
D2 i2
−VZ
V
Von V1 (or V2 )
* i > 0 → D1 in forward conduction, D2 in reverse conduction
→ V1 = Von , V2 = −VZ .
Total voltage drop V = V1 − V2 = Von + VZ .
Example: Von = 0.7 V, VZ = 5 V → V = 5.7 V.
* i < 0 → D1 in reverse conduction, D2 in forward conduction
→ V1 = −VZ , V2 = Von .
M. B. Patil, IIT Bombay
Two Zener diodes connected “back-to-back”
i1 (or i2 )
V1
i1 D1
i
V2
D2 i2
−VZ
V
Von V1 (or V2 )
* i > 0 → D1 in forward conduction, D2 in reverse conduction
→ V1 = Von , V2 = −VZ .
Total voltage drop V = V1 − V2 = Von + VZ .
Example: Von = 0.7 V, VZ = 5 V → V = 5.7 V.
* i < 0 → D1 in reverse conduction, D2 in forward conduction
→ V1 = −VZ , V2 = Von .
Total voltage drop V = V1 − V2 = −VZ − Von = −(VZ + Von ) = −5.7 V.
M. B. Patil, IIT Bombay
Two Zener diodes connected “back-to-back”
i1 (or i2 )
V1
i1 D1
i
V2
D2 i2
−VZ
V
Von V1 (or V2 )
* i > 0 → D1 in forward conduction, D2 in reverse conduction
→ V1 = Von , V2 = −VZ .
Total voltage drop V = V1 − V2 = Von + VZ .
Example: Von = 0.7 V, VZ = 5 V → V = 5.7 V.
* i < 0 → D1 in reverse conduction, D2 in forward conduction
→ V1 = −VZ , V2 = Von .
Total voltage drop V = V1 − V2 = −VZ − Von = −(VZ + Von ) = −5.7 V.
* For −(VZ + Von ) < V < (VZ + Von ), conduction is not possible → i = 0.
M. B. Patil, IIT Bombay
Two Zener diodes connected “back-to-back”
i1 (or i2 )
V1
i1 D1
i
V2
D2 i2
−VZ
Von V1 (or V2 )
V
* i > 0 → D1 in forward conduction, D2 in reverse conduction
→ V1 = Von , V2 = −VZ .
Total voltage drop V = V1 − V2 = Von + VZ .
Example: Von = 0.7 V, VZ = 5 V → V = 5.7 V.
* i < 0 → D1 in reverse conduction, D2 in forward conduction
→ V1 = −VZ , V2 = Von .
Total voltage drop V = V1 − V2 = −VZ − Von = −(VZ + Von ) = −5.7 V.
* For −(VZ + Von ) < V < (VZ + Von ), conduction is not possible → i = 0.
i
−(VZ + Von )
V
(VZ + Von )
M. B. Patil, IIT Bombay
Diode circuit example (voltage limiter)
R
1k
i
D1
Vi
Vo
D2
Von = 0.7 V, VZ = 5 V.
Plot Vo versus Vi .
Diode circuit example (voltage limiter)
R
1k
i
i
D1
Vi
−(VZ + Von )
Vo
D2
Von = 0.7 V, VZ = 5 V.
Plot Vo versus Vi .
Vo
(VZ + Von )
Diode circuit example (voltage limiter)
R
1k
i
i
D1
Vi
−(VZ + Von )
Vo
D2
Vo
(VZ + Von )
Von = 0.7 V, VZ = 5 V.
Plot Vo versus Vi .
* For −5.7 V < Vi < 5.7 V, no conduction is possible → Vo = Vi .
Diode circuit example (voltage limiter)
R
8
1k
i
i
D1
Vi
4
−(VZ + Von )
Vo
D2
Vo
(VZ + Von )
Vo
0
−4
Von = 0.7 V, VZ = 5 V.
Plot Vo versus Vi .
* For −5.7 V < Vi < 5.7 V, no conduction is possible → Vo = Vi .
−8
−8
−4
0
4
8
Vi
Diode circuit example (voltage limiter)
R
8
1k
i
i
D1
Vi
4
−(VZ + Von )
Vo
D2
Vo
(VZ + Von )
Vo
0
−4
Von = 0.7 V, VZ = 5 V.
Plot Vo versus Vi .
−8
−8
−4
0
4
* For −5.7 V < Vi < 5.7 V, no conduction is possible → Vo = Vi .
* For Vi > 5.7 V, D1 is forward-biased, D2 is reverse-biased, and Vo = (Von + VZ ). The excess voltage
(Vi − (Von + VZ )) drops across R.
8
Vi
Diode circuit example (voltage limiter)
R
8
1k
i
i
D1
Vi
4
−(VZ + Von )
Vo
D2
Vo
(VZ + Von )
Vo
0
−4
Von = 0.7 V, VZ = 5 V.
Plot Vo versus Vi .
−8
−8
−4
0
4
* For −5.7 V < Vi < 5.7 V, no conduction is possible → Vo = Vi .
* For Vi > 5.7 V, D1 is forward-biased, D2 is reverse-biased, and Vo = (Von + VZ ). The excess voltage
(Vi − (Von + VZ )) drops across R.
8
Vi
Diode circuit example (voltage limiter)
R
8
1k
i
i
D1
Vi
4
−(VZ + Von )
Vo
D2
Vo
(VZ + Von )
Vo
0
−4
Von = 0.7 V, VZ = 5 V.
Plot Vo versus Vi .
−8
−8
−4
0
4
* For −5.7 V < Vi < 5.7 V, no conduction is possible → Vo = Vi .
* For Vi > 5.7 V, D1 is forward-biased, D2 is reverse-biased, and Vo = (Von + VZ ). The excess voltage
(Vi − (Von + VZ )) drops across R.
* For Vi < −5.7 V, D2 is forward-biased, D1 is reverse-biased, and Vo = −(Von + VZ ). The excess voltage
(−Vi − (Von + VZ )) drops across R.
8
Vi
Diode circuit example (voltage limiter)
R
8
1k
i
i
D1
Vi
4
−(VZ + Von )
Vo
D2
Vo
(VZ + Von )
Vo
0
−4
Von = 0.7 V, VZ = 5 V.
Plot Vo versus Vi .
−8
−8
−4
0
4
8
Vi
* For −5.7 V < Vi < 5.7 V, no conduction is possible → Vo = Vi .
* For Vi > 5.7 V, D1 is forward-biased, D2 is reverse-biased, and Vo = (Von + VZ ). The excess voltage
(Vi − (Von + VZ )) drops across R.
* For Vi < −5.7 V, D2 is forward-biased, D1 is reverse-biased, and Vo = −(Von + VZ ). The excess voltage
(−Vi − (Von + VZ )) drops across R.
M. B. Patil, IIT Bombay